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maw [93]
3 years ago
10

Which of the following is a large-scale, slow-moving natural change?

Chemistry
1 answer:
KIM [24]3 years ago
7 0
A) glaciation because the block of ice slowly moves through the ocean and can destroy the icy habitat by its strong weight.
You might be interested in
I need help solving this!
zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
)
xenn [34]

Answer:

Its C hope it helped

Explanation:

3 0
3 years ago
Can a molecule have polar bonds, but still be a no polar molecule?
My name is Ann [436]

<em><u>A molecule </u></em><em><u>can </u></em><em><u>possess polar bonds and still be nonpolar.</u></em>

I hope this helped. Have a nice day, make sure to take care of yourself. You're loved <3

5 0
3 years ago
Read 2 more answers
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
1 year ago
Oxygen-Carbon Dioxide Exchange
leonid [27]

Answer:

Answer is below with the steps in order

Explanation:

4) Blood picks up carbon dioxide from the body

7) Heart pumps carbon dioxide rich blood to the lungs

8) Inhale

9) Nose traps germs in air

10) Air moves down the trachea

3) Air moves through bronchi into the bronchioles

6) Alveoli receive oxygen pass to blood

9) Oxygen passes into the blood

11) Alveoli receives carbon dioxide from the blood

12) Oxygen-rcih blood flows to the heart

3) Carbon Dioxide moves from bonchioles to bronchi

13)Carbon dioxide flows up

1) Heart pumps oxygen-rich blood to the body

The last <em>four of five</em> steps could be switched up a little bit since it all happens synchronously, but this is the most accurate interpretation.

8 0
2 years ago
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